How do you find the Vertical, Horizontal, and Oblique Asymptote given #f(x)= (2x+1)/(x-1)#?

Question

Serenity Jones · Accepted Answer

vertical asymptote at x = 1
horizontal asymptote at y = 2

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote. solve: #x-1=0rArrx=1" is the asymptote"# Horizontal asymptotes occur as #lim_(xto+-oo),f(x)toc" (a constant)"# divide terms on numerator/denominator by x #f(x)=((2x)/x+1/x)/(x/x-1/x)=(2+1/x)/(1-1/x)# as #xto+-oo,f(x)to(2+0)/(1-0)# #rArry=2" is the asymptote"# Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (both of degree 1 ) Hence there are no oblique asymptotes. graph{(2x+1)/(x-1) [-20, 20, -10, 10]}

Serenity Jones · Answer

undefined To find the vertical asymptote(s) of a function, determine the values of $ x $ that make the denominator equal to zero. In this case, the denominator is $ x - 1 $, so the vertical asymptote occurs at $ x = 1 $.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator of the function. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is at $ y = 0 $. If the degrees are equal, divide the leading coefficient of the numerator by the leading coefficient of the denominator to find the horizontal asymptote. Here, the degrees are equal, so the horizontal asymptote is $ y = 2/1 = 2 $.

There are no oblique asymptotes because the degree of the numerator is not greater than the degree of the denominator.